Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
using namespace std;
double D[3010][3010];
int n;
int main() {
	int i, j;
	D[0][0] = 1;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		double p;
		scanf("%lf", &p);
		for (j = 0; j < i; j++) {
			if (D[i - 1][j] < 1e-40)D[i - 1][j] = 0;
			D[i][j + 1] += D[i - 1][j] * p;
			D[i][j] += D[i - 1][j] * (1-p);
		}
	}
	double res = 0.0;
	for (i = n / 2 + 1; i <= n; i++)res += D[n][i];
	printf("%.15f\n", res);
}